Back to the on-screen lesson ·
Complex numbers as points of a plane: addition component by component, a multiplication that mixes the components, the conjugate that makes a denominator real, and multiplication by $i$ as a quarter turn.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to add, subtract, multiply and divide complex numbers written in the form $x + iy$, use the conjugate to make a denominator real, read off the real and imaginary parts of a result, simplify any power of the imaginary unit, and say why multiplication in the plane is a rotation together with a stretching rather than anything a vector does.
Arithmetic with square roots of negative numbers, from the quadratic formula: the symbol $i$ with $i^{2} = -1$, and expressions of the form $x + iy$. What is new here is the insistence that $x + iy$ is a point of a plane, because every later theorem in this course is a statement about that plane rather than about a formula.
For $z = x + iy$ with $x$ and $y$ real, $x$ is the real part $\operatorname{Re} z$ and $y$ the imaginary part $\operatorname{Im} z$ — a real number, despite the name. The conjugate $\bar z$ is $x - iy$, the reflection of $z$ in the real axis. A number with $x = 0$ is purely imaginary; one with $y = 0$ is real. The plane in which $z$ is drawn is the complex plane, and $\mathbb{C}$ names the set of all such numbers.
A complex number is a point $z = x + iy$ of the plane, and the whole subject rests on the two operations it carries.
Addition is the plane's. $(x + iy) + (u + iv) = (x + u) + i(y + v)$: components add, exactly as vectors do, and the picture is the parallelogram rule.
Multiplication is new. Expand the brackets and use $i^{2} = -1$:
$$(x + iy)(u + iv) = (xu - yv) + i(xv + yu).$$
Nothing in vector algebra does this. The two components are mixed, and one term crosses from the imaginary side to the real side with its sign reversed. That single crossing is what makes $\mathbb{C}$ a field rather than merely a plane: every non-zero number has a multiplicative inverse.
The conjugate is the tool that makes division work. $\bar z = x - iy$ satisfies
$$z\bar z = x^{2} + y^{2},$$
a real number, zero only when $z$ is. So
$$\frac{1}{z} = \frac{\bar z}{z \bar z} = \frac{x - iy}{x^{2} + y^{2}},$$
and every quotient is computed by multiplying top and bottom by the conjugate of the bottom. Conjugation also respects both operations: $\overline{z + w} = \bar z + \bar w$ and $\overline{zw} = \bar z\,\bar w$.
And $i$ is a quarter turn. Multiplying by $i$ sends $x + iy$ to $-y + ix$, which is the point rotated a quarter turn anticlockwise about the origin. Four of those is the identity, which is the whole reason the powers of $i$ repeat with period four.
Another way: picture
Draw $1$ on the real axis and multiply by $i$ four times: the point walks round the corners of a square — $1$, $i$, $-1$, $-i$ — and returns. Multiplication by a complex number is a rotation together with a stretching, and every later fact about powers, roots and the exponential is a consequence of that one observation.
Another way: steps
Conjugation looks like a piece of notation and is in fact the course's first real tool. Three uses, all of which return later.
It makes things real. $z + \bar z = 2x$ and $z - \bar z = 2iy$, so $\operatorname{Re} z = \tfrac12(z + \bar z)$ and $\operatorname{Im} z = \tfrac{1}{2i}(z - \bar z)$. Anything that can be said about real and imaginary parts can be said about $z$ and $\bar z$ instead, and that is the form in which the Cauchy-Riemann equations will eventually be stated most cleanly.
It makes denominators real. That is division, above.
It is the one operation of the plane that complex differentiation cannot see. This is the fact to remember. Everything in this course that is built from $z$ alone will turn out to be beautifully behaved, and everything that needs $\bar z$ — the conjugate itself, $\operatorname{Re} z$, $\operatorname{Im} z$, $|z|$ — will turn out to be analytic nowhere. Notice which of those four is written with a conjugate and you have predicted most of unit 2.
The most expensive habit at this stage is treating $x + iy$ as the vector $(x, y)$ and multiplication as something bolted on. Two things go wrong.
A product of two imaginary numbers is real. $(2i)(3i) = -6$, not $6i$ and not a vector at all. Every term with two factors of $i$ crosses to the real part with its sign reversed, and forgetting the sign is the single commonest arithmetic error in the whole course.
There is no order. Real numbers can be compared and complex ones cannot: $i > 0$ and $i < 0$ are both false, and no sensible ordering exists. A statement like the larger root means nothing here. What can be compared is the modulus, which is a real number, and that is what the next lesson is about.
Complex differentiability is not the real kind with a letter added. The difference quotient has to settle down to the same number along every one of the infinitely many directions a point can be approached from in a plane, and that single demand is strong enough to force a function to have derivatives of every order, to equal its own Taylor series, and to be determined on a whole region by its values on a curve. Nothing in real calculus behaves like that, so a fact carried over from it without checking is a guess.
$(3 + 2i)(4 - 5i)$: the four terms are $12$, $-15i$, $8i$ and $-10i^{2}$.
Expand as ordinary algebra first.
$-10i^{2} = +10$, so the real part is $12 + 10 = 22$ and the imaginary part is $-15 + 8 = -7$: the answer is $22 - 7i$.
The square of the imaginary unit moves a term across.
$\dfrac{2 + i}{3 - i}$: multiply top and bottom by $3 + i$.
The conjugate of the denominator.
The bottom becomes $9 + 1 = 10$; the top becomes $6 + 2i + 3i - 1 = 5 + 5i$.
A real denominator at last.
So the quotient is $\tfrac{1}{2} + \tfrac{1}{2}i$.
Divide each part by the real number.
The conjugate of the denominator is $1 + 2i$, and $(1 - 2i)(1 + 2i) = 1 + 4 = 5$.
Make the denominator real.
So the reciprocal is $\tfrac{1}{5} + \tfrac{2}{5}i$.
Let $z = 7 + 5i$ and $w = 7 + 7i$. Fill in the table.
| Value | |
|---|---|
| The real part of the sum | |
| The imaginary part of the sum | |
| The real part of the product | |
| The imaginary part of the product |
What is the imaginary part of $(6 + 8i)(4 + 5i)$?
Answer:
Match each power of $i$ to the number it equals.
| $1$ | $i$ | $-1$ | $-i$ | |
|---|---|---|---|---|
| $i^{16}$ | ||||
| $i^{17}$ | ||||
| $i^{18}$ | ||||
| $i^{19}$ |
Put the steps of writing $\dfrac{1}{2 + 3i}$ in the form $x + iy$ in order.
Number the steps in order (write the number in the box):
What kind of number is $(9 + 6i)(9 - 6i)$?
Add together $i + i^{2} + i^{3} + \cdots + i^{24}$. What do you get?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $z = 5 + 8i$ and $w = 1 + 7i$. Fill in the table.
| Value | |
|---|---|
| The real part of the sum | |
| The imaginary part of the sum | |
| The real part of the product | |
| The imaginary part of the product |
You can compute with complex numbers in rectangular form and you can say what the conjugate is for. Say in your own words why a number times its conjugate is never negative. Next: the modulus, which turns that product into a distance.
9. Your turn: write the reciprocal of $1 - 2i$ in the form $x + iy$, step 2